Source: José Figueroa-O'Farrill's Mathematical Techniques III Lecture Notes
Trying to evaluate the following integral:
$\oint_{\text{ }\Gamma}\frac{2z+1}{z(z-1)^2}dz$
where the contour $\Gamma$ is given by
By Cauchy's Integral Theorem, $\Gamma$ is equivalent to the pair of contours $\Gamma_0$ and $\Gamma_1$, given by
The integrand can be brought to the following form via partial fraction decomposition,
$\frac{}{}=\frac{3}{(z-1)^2} -\frac{1}{z-1} +\frac{1}{z}$
Now, can someone explain to me when the contour integrals about $\Gamma_0$ and $\Gamma_1$ respectively are only non-zero for the following terms in the above decomposition?
It must be from Cauchy's Integral Theorem, so maybe I really don't understand that theorem.
Consider, for instance, $\frac1z$, whose domain is $\mathbb C\setminus\{0\}$. Then the loop $\Gamma_1$ is homotopically null (that is, it is homotopic to a constant loop) in $\mathbb C\setminus\{0\}$, and therefore $\int_{\Gamma_1}\frac{\mathrm dz}z=0$. So$$\int_{\Gamma_1}\frac{\mathrm dz}{(z-1)^2}+\int_{\Gamma_1}-\frac{\mathrm dz}{z-1}+\int_{\Gamma_1}\frac{\mathrm dz}z=\int_{\Gamma_1}\frac{\mathrm dz}{(z-1)^2}+\int_{\Gamma_1}-\frac{\mathrm dz}{z-1}.$$